Problem: The polynomial $p(x)=5x^3-9x^2-6x+8$ has a known factor of $(x+1)$. Rewrite $p(x)$ as a product of linear factors. $p(x)=$
Explanation: We know $(x+1)$ is a factor of $p(x)$. This means that $p(x)=(x+1)\cdot q(x)$ for some polynomial $q(x)$. We can find $q(x)$ using polynomial division, and then we can factor $q(x)$. This way, we will be able to rewrite $p(x)$ as a product of linear factors. Dividing $p(x)$ by $(x+1)$ $\begin{array}{r} 5x^2-14x+8 \\ x+1|\overline{5x^3-\phantom{1}9x^2-\phantom{1}6x+8} \\ \mathllap{-(}\underline{5x^3+\phantom{1}5x^2\phantom{-16x+8}\rlap )} \\ -14x^2-\phantom{1}6x+8 \\ \mathllap{-(}\underline{-14x^2-14x\phantom{+8}\rlap )} \\ 8x+8 \\ \mathllap{-(}\underline{8x+8\rlap )} \\ 0 \end{array}$ We find that $q(x)=5x^2-14x+8$. Factoring $q(x)$ We can factor $q(x)$ by grouping: $\begin{aligned} q(x)&=5x^2-14x+8 \\\\ &=5x^2-10x-4x+8 \\\\ &=5x(x-2)-4(x-2) \\\\ &=(5x-4)(x-2) \end{aligned}$ Putting it all together $\begin{aligned} p(x)&=5x^3-9x^2-6x+8 \\\\ &=(x+1)(5x^2-14x+8) \\\\ &=(x+1)(5x-4)(x-2) \end{aligned}$